Difference between revisions of "1952 AHSME Problems/Problem 49"
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== Solution == | == Solution == | ||
Let <math>[ABC]=K.</math> Then <math>[ADC] = \frac{1}{3}K,</math> and hence <math>[N_1DC] = \frac{1}{7} [ADC] = \frac{1}{21}K.</math> Similarly, <math>[N_2EA]=[N_3FB] = \frac{1}{21}K.</math> Then <math>[N_2N_1CE] = [ADC] - [N_1DC]-[N_2EA] = \frac{5}{21}K,</math> and same for the other quadrilaterals. Then <math>[N_1N_2N_3]</math> is just <math>[ABC]</math> minus all the other regions we just computed. That is, <cmath>[N_1N_2N_3] = K - 3\left(\frac{1}{21}K\right) - 3\left(\frac{5}{21}\right)K = K - \frac{6}{7}K = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> | Let <math>[ABC]=K.</math> Then <math>[ADC] = \frac{1}{3}K,</math> and hence <math>[N_1DC] = \frac{1}{7} [ADC] = \frac{1}{21}K.</math> Similarly, <math>[N_2EA]=[N_3FB] = \frac{1}{21}K.</math> Then <math>[N_2N_1CE] = [ADC] - [N_1DC]-[N_2EA] = \frac{5}{21}K,</math> and same for the other quadrilaterals. Then <math>[N_1N_2N_3]</math> is just <math>[ABC]</math> minus all the other regions we just computed. That is, <cmath>[N_1N_2N_3] = K - 3\left(\frac{1}{21}K\right) - 3\left(\frac{5}{21}\right)K = K - \frac{6}{7}K = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> | ||
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+ | == Solution 2 == | ||
+ | We can force this triangle to be equilateral because the ratios are always <math>3:3:1</math>, and with the symmetry of the equilateral triangle, it is safe to assume that this may be the truth. Then, we can do a simple coordinate bash. Let <math>B</math> be at <math>(0,0)</math>, <math>A</math> be at <math>(3,3\sqrt{3})</math>, and <math>C</math> be at <math>(6,0)</math>. We then create a new point <math>O</math> at the center of everything. It should be noted because of similarity between <math>\triangle N_{1}N_{2}N_{3}</math> and <math>\triangle ABC</math>, we can find the scale factor between the two triangle by simply dividing <math>AO</math> by <math>N_{2}O</math> (nitrous oxide). First, we need to find the coordinates of <math>O</math> and <math>N_{2}</math>. <math>O</math> is easily found at <math>(3,\sqrt{3})</math> and <math>N_{2}</math> be found by calculating equation of <math>BE</math> and <math>AD</math>.<math>E</math> is located <math>(4,2\sqrt{3})</math> so <math>BE</math> is <math>y=\frac{x\sqrt{3}}{2}</math>. <math>D</math> be at <math>(4,0)</math> and the slope is <math>-3\sqrt{3}</math>. We see that they be at the same <math>x</math>-value. Quick maths calculate the x value to be <math>4-\frac{2\sqrt{3}}{3\sqrt{3}+\frac{\sqrt{3}}{2}}</math> which be <math>3\frac{3}{7}</math>. Another quick maths caculation of the <math>y</math>-value lead it be equal <math>2\sqrt{3}*\frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}+3\sqrt{3}}</math> which be <math>1\frac{5}{7}\sqrt{3}</math>. Peferct, so now <math>N_{2}</math> be at <math>(3\frac{3}{7},1\frac{5}{7}\sqrt{3})</math>. Subtracting the coordinate with the center give you <math>(\frac{3}{7}, \frac{5}{7}\sqrt{3})</math>. I don't even want to do this anymore so here is the answer: <cmath>\boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> | ||
== See also == | == See also == |
Revision as of 15:59, 6 November 2020
Contents
[hide]Problem
In the figure, , and are one-third of their respective sides. It follows that , and similarly for lines BE and CF. Then the area of triangle is:
Solution
Let Then and hence Similarly, Then and same for the other quadrilaterals. Then is just minus all the other regions we just computed. That is,
Solution 2
We can force this triangle to be equilateral because the ratios are always , and with the symmetry of the equilateral triangle, it is safe to assume that this may be the truth. Then, we can do a simple coordinate bash. Let be at , be at , and be at . We then create a new point at the center of everything. It should be noted because of similarity between and , we can find the scale factor between the two triangle by simply dividing by (nitrous oxide). First, we need to find the coordinates of and . is easily found at and be found by calculating equation of and . is located so is . be at and the slope is . We see that they be at the same -value. Quick maths calculate the x value to be which be . Another quick maths caculation of the -value lead it be equal which be . Peferct, so now be at . Subtracting the coordinate with the center give you . I don't even want to do this anymore so here is the answer:
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
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All AHSME Problems and Solutions |
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